Home
Class 12
CHEMISTRY
In the reduction of nitrous oxide with...

In the reduction of nitrous oxide with hydrogen an equimolar mixture of the gases at 340 mm initial pressure was half changed in 120 second in a second experiment when the intial pressure 288 mm the change was half completed in 140 seconds what is the order of reaction

Text Solution

Verified by Experts

Since this is a gaseous reaction, concentrations can be substituted by pressures.
Thus, `(t_(1))/(t_(2)) = ((a_(2))/(a_(1)))^(n-1) = [(p_(2))/(p_(1))]^(n-1)`
Given : `p_(1) = 340 mm, p_(2) = 288 mm, t_(1) = 102s, t_(2) = 140s`
`therefore (102)/(140) = [(288)/(340)]^(n-1)` or `(n-1) ln""(288)/(340) = ln ""(102)/(140), n = 2.9 ~~3`
The reaction is of the third order.
Promotional Banner

Similar Questions

Explore conceptually related problems

0.75 mole of solid A_(4) and 2 moles of gaseous O_(2) are heated in a sealed vessel, completely using up the reactants and producing only one compound. It is found that when the temperature is reduced to the initial temperature, the contents of the vessel exhibit a pressure equal to half the original pressure. What conclusions can be drawn from these data about the product of the reaction?

For the reaction system: 2NO(g) + O_(2) (g) rarr 2NO_(2) (g) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first-order with respect to O_(2) and second-order with respect to NO, the rate of reaction will be

The reaction 2NO_(2) rarr2NO+O_(2) is second order respect to NO_(2) . If the intial concentration of NO_(2)(g) is 6.54 xx10^(-4) mol L^(-1) and the intial reaction rate is 4.42 xx10^(-7) mol L^(-1) s^(-1) what is the half life of the reaction.